Let $R$ be the region enclosed by the polar curve $r(\theta)=2\theta\sin(\theta)$ where $\dfrac{\pi}{3}\leq \theta\leq \dfrac{2\pi}{3}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}2\theta^2\sin^2(\theta)\,d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{2\pi}{3}}2\theta^2\sin^2(\theta)\,d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{2\pi}{3}}\theta\sin(\theta)\,d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}\theta\sin(\theta)\,d\theta$
Solution: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=2\theta\sin(\theta)}$, ${\alpha=\dfrac{\pi}{3}}$, and ${\beta=\dfrac{2\pi}{3}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{3}}}^{{\scriptsize\dfrac{2\pi}{3}}}\dfrac{1}{2}\left({2\theta\sin(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{2\pi}{3}}\dfrac{1}{2}\cdot4\theta^2\sin^2(\theta)\,d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{2\pi}{3}}2\theta^2\sin^2(\theta)\,d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{3}}^{\scriptsize\dfrac{2\pi}{3}}2\theta^2\sin^2(\theta)\,d\theta$